Final answer:
When two charged conducting spheres touch, their charges redistribute evenly. The potential energy of the system decreases in this process, making (C) always a decrease in energy of the system the correct answer.
Step-by-step explanation:
When considering two conducting spheres of radii R1 and R2 charged with charges Q1 and Q2 respectively, bringing them into contact allows the charges to redistribute over the surface of both spheres due to the principle of conserving charge.
The final charge on each sphere, after they have touched and been separated, will be the same as they share the total original charge evenly. When the spheres are initially charged, they each have an electric potential energy that is dependent on their charge and radius. According to the formula V = kQ/R, where V is the electric potential, k is Coulomb's constant, Q is the charge, and R is the radius of the sphere. We can use this to understand that the electric potentials of the spheres are equal when they have been connected by a wire and share a common charge.
The electrical potential energy of the system can be considered positive if the charges are alike and negative if the charges are opposite. Bringing identical charges closer together requires one to do work against the electrostatic repulsive force, which results in an increase in potential energy. Conversely, bringing opposite charges closer together allows the system to reduce its potential energy because they naturally attract each other. Hence, when we bring two charged objects into contact, we are rearranging the charges in a way that affects the system's energy.
Considering the information given, option (C) always a decrease in energy of the system would be the correct answer because when the spheres are brought together and charges are redistributed, the potential energy of the system decreases due to charge redistribution over a larger combined surface.